Strong Acid-Strong Base Titrations
Stoichiometric equivalence and excess-reagent pH
Lesson 1811 of 4,500 · Equilibrium: Chemical and Ionic
Learning objectives
- Calculate pH before and after equivalence
- Explain pH at strong acid-strong base equivalence
Introduction
Strong acid-strong base titration is primarily a mole-accounting problem. Hydronium and hydroxide neutralize nearly completely, so the excess strong reagent controls pH away from equivalence. Near equivalence the difference between large opposing mole totals becomes small, producing a steep curve. At 25 °C, a simple salt of a strong acid and strong base gives nearly neutral equivalence.
Core explanation
For HCl titrated with NaOH, the net ionic reaction is H₃O⁺ + OH⁻ → 2H₂O. Before equivalence, n(H₃O⁺ initial) exceeds n(OH⁻ added); subtract to find remaining acid moles. Divide by combined solution volume to obtain hydronium concentration, then pH ≈ −log₁₀[H₃O⁺] for ordinary dilute calculations. After equivalence, subtract initial acid moles from added hydroxide moles, divide by total volume, calculate pOH, and use pH + pOH ≈ 14.00 at 25 °C.
At exact equivalence, neither strong reagent remains in excess. Sodium and chloride are spectators to the pH calculation in this simple system. Water autoionization sets pH near 7.00 at 25 °C. Neutral pH is not always numerically 7 because Kw varies with temperature. Also, strongly concentrated solutions require activity considerations; the classroom pH equations are dilute-solution approximations.
The equivalence volume follows stoichiometry. For a monoprotic strong acid and a base providing one OH⁻ per formula unit, CaVa = CbVeq. If a base provides two OH⁻ equivalents, the factor must appear in the balanced equation. Do not let the shortcut hide reaction coefficients. Precision of titrant concentration and delivered volume determines precision of the analyte result.
The steep jump is not a discontinuity. In the vicinity of equivalence, the excess-reagent approximation becomes weak because water autoionization can matter. A full calculation can use water equilibrium plus charge balance. In most school problems, exact equivalence is handled separately and nearby points are chosen far enough away that leftover strong reagent dominates. The pH curve's shape also depends on sample concentration: more dilute solutions show a less dramatic jump.
Step-by-step reasoning
1. Compute initial acid and delivered base moles using actual volumes. 2. Subtract according to the net ionic reaction. 3. Divide excess moles by total mixed volume. 4. Use pH for acid excess or pOH then pH for base excess.
Visual explanation
Draw a mole line with acid remaining to the left of equivalence and hydroxide remaining to the right. At the center, label the strong-reagent difference zero.
Real-world analogy
Two teams cancel matching tokens one for one. Before one team's pile runs out, only its leftovers matter; after that crossing, the other team's leftovers determine the count.
Real-world example
Standardized sodium hydroxide can determine the amount of a strong acid in an unknown sample. The endpoint is most reliable when the indicator transition lies within the steep pH rise.
Why?
Why does a simple HCl/NaOH titration have pH near 7 at equivalence at 25 °C? Neither strong reagent remains, and the Na⁺ and Cl⁻ ions do not appreciably hydrolyze water.
Common misconception
“At half the equivalence volume, pH equals pKa.” That half-equivalence rule applies to an appropriate weak-acid buffer region, not to a strong acid.
Worked example
Mix 25.0 mL of 0.100 M HCl with 20.0 mL of 0.100 M NaOH. Initial acid = 0.00250 mol; added base = 0.00200 mol. Acid excess = 0.00050 mol in 0.0450 L, so [H₃O⁺] ≈ 0.0111 M and pH ≈ 1.95. If 30.0 mL base had been added instead, OH⁻ excess would be 0.00050 mol in 0.0550 L; pOH ≈ 2.04 and pH ≈ 11.96 at 25 °C.
Quick check
1. Which concentration controls pH after excess NaOH has been added? Answer: The leftover OH⁻ concentration after neutralization and dilution.
Exam focus
Always add acid and base volumes before dividing leftover moles by volume. At exact equivalence, do not calculate pH from zero excess moles with a logarithm.
Advanced insight
Near exact equivalence, water contributes appreciable hydronium and hydroxide compared with a tiny strong-reagent difference. Charge balance with Kw provides a smooth curve through this region.
Summary
Strong acid-strong base titration uses neutralization stoichiometry first. Excess hydronium controls pH before equivalence; excess hydroxide controls it afterward. At simple equivalence, water controls pH.
Practice questions
1. What is the equivalence volume for 20.0 mL of 0.100 M HCl titrated by 0.100 M NaOH? Answer: 20.0 mL, because both deliver 0.00200 mol equivalents. 2. Why is total mixed volume needed after neutralization? Answer: Leftover moles must be divided by the final solution volume to give concentration. 3. Does neutral always mean pH 7.00 at every temperature? Answer: No. Neutrality means equal hydronium and hydroxide activities; Kw changes with temperature.